When adding extra electrons to pure silicon in a conduction band, the extra electrons will occupy a donor level below the conduction band (we get n-type), as shown here:enter image description here

When adding electron-deficient dopant atom to pure silicon in a valence band, this provides an acceptor level above the valence band (we get p-type), as shown here:enter image description here

Is this definition of doping semiconductors true? Is this donor energy part of conduction band or it is between conduction and valence band (so in forbidden gap)? And same for acceptor energy?


2 Answers 2


The diagrams appear correct - although as with all diagrams of this sort they don't quite tell the full story, but good enough for a basic understanding.

When you add the dopants in, they add extra energy levels within the gap between valence and conduction bands that electrons can sit at Reference. These energy levels represent the ionisation energy of the dopant.

When the device is very cold (e.g. if you put it in LN2), there is not enough energy to allow the electrons to move either up to acceptor energy levels, or up in to the conduction band from donor energy levels.

However, as soon as you warm the device up, there is enough energy to ionise the dopants. For p-type, electrons gain enough thermal energy to hop from the valence band into the energy states created by the p-type dopant ions. For n-type, the electrons gain enough energy to escape the donor ions and move in to the conduction band.

It is important to note though that the diagrams don't give a good idea of scale - the energy difference between the band edge and dopant levels are much smaller than the band gap of the semiconductor. The dopant levels might be only 0.05eV Reference from the band edge compared to a band gap of say 0.7eV Reference. As a result the excited electrons from the donors are far more likely to go in to the conduction band than the valence band, and the acceptors are far more likely to pull in electrons from the valence band.

  • \$\begingroup\$ Where did the “0.05 eV” figure come from? \$\endgroup\$ Sep 3, 2016 at 19:18
  • \$\begingroup\$ @IncnisMrsi that was just from an example that I had in my Uni notes, it's not necessarily a typical figure, but it will be much less than the band gap. \$\endgroup\$ Sep 3, 2016 at 19:36
  • \$\begingroup\$ Is the concept of dopant band also from “your Uni notes”? Could you elaborate, how is it a band distinct from conduction(n) or valence(p)? \$\endgroup\$ Sep 3, 2016 at 19:48
  • \$\begingroup\$ @IncnisMrsi my bad, that was supposed to say dopant levels (as it says is a couple of other places in the answer). Fixed. \$\endgroup\$ Sep 3, 2016 at 19:56

There is no such thing as “dopant band” distinct from regular semiconductor’s bands. Dopant’s atoms might create some bound electron states. At zero temperature, they are either filled (for n-type doping) or vacant (for p-type doping). This is the thing possibly depicted at the OP’s picture. But they don’t play a great rôle at normal temperatures, when a doped semiconductor operates as a doped semiconductor.

In a n-type semiconductor “extra electrons” are in the conduction band, near its bottom. Such semiconductor type has more electrons than a pure material. The conduction band is (practically) the same.

In a p-type semiconductor holes are in the valence band, near its top. Such semiconductor type has less electrons than a pure material, hence the valence band is not entirely occupied. The valence band is (practically) the same.

All the above is for the case where the semiconductor is only lightly doped.

  • \$\begingroup\$ When the temperature is sufficient, I totally agree, the carriers are excited into the conduction (n-type) and valence (p-type) bands. My answer does not say otherwise. However, when you cool the device down, it returns to acting as an intrinsic semiconductor, and the carriers return to the dopants. I'm pretty sure I am not imagining what is in all the diagrams I've seen. \$\endgroup\$ Sep 3, 2016 at 19:45
  • \$\begingroup\$ @Tom Carpenter: Ī̲ had misconceptions. Thank you for feedback. \$\endgroup\$ Sep 3, 2016 at 20:25

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